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Chapter 13: Problem 23

Deimos, a moon of Mars, is about 12 \(\mathrm{km}\) in diameter with mass \(2.0 \times 10^{15} \mathrm{kg} .\) Suppose you are stranded alone on Deimos and want to play a one-person game of baseball. You would be the pitcher, and you would be the batter! (a) With what speed would you have to throw a baseball so that it would go into a circular orbit just above the surface and return to you so you could hit it? Do you think you could actually throw it at this speed? (b) How long (in hours) after throwing the ball should you be ready to hit it? Would this be an action-packed baseball game?

Short Answer

Expert verified
To orbit Deimos, throw at 5.6 m/s. Ball returns in 1.87 hours. Not action-packed.

Step by step solution

01

Identify the Formula for Orbital Speed

To find the speed needed to put the baseball in a circular orbit, we use the formula for orbital velocity: \( v = \sqrt{\frac{GM}{R}} \).Here, \( G \) is the gravitational constant \( (6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2) \), \( M \) is the mass of Deimos, and \( R \) is the radius of Deimos.
02

Calculate the Radius of Deimos

The radius of Deimos \( R \) is half its diameter. Given that the diameter is \( 12 \, \text{km} \), \( R = 6 \, \text{km} = 6,000 \, \text{m} \).
03

Use the Mass of Deimos

The mass of Deimos \( M \) is given as \( 2.0 \times 10^{15} \, \text{kg} \).
04

Calculate the Orbital Velocity

Substitute \( G \), \( M \), and \( R \) into the orbital velocity formula:\[ v = \sqrt{\frac{(6.674 \times 10^{-11}) \times (2.0 \times 10^{15})}{6,000}} \approx 5.6 \, \text{m/s} \].This is the speed required for the baseball to orbit Deimos just above the surface.
05

Determine if You Could Throw at this Speed

Consider that a professional baseball pitcher can throw a baseball at about 40 m/s, which is much higher than 5.6 m/s. Thus, it is possible to throw a baseball at this speed.
06

Calculate the Orbital Period

The time it takes for the baseball to complete one orbit (orbital period \( T \)) is found using the formula \( T = \frac{2\pi R}{v} \).
07

Plug in the Values to Find the Orbital Period

Substitute \( R = 6,000 \, \text{m} \) and \( v = 5.6 \, \text{m/s} \) into the formula:\[ T = \frac{2\pi \times 6,000}{5.6} \approx 6,734 \, \text{s} \].
08

Convert the Time to Hours

Convert seconds to hours:\[ 6,734 \, \text{s} \times \frac{1 \, \text{hour}}{3,600 \, \text{s}} \approx 1.87 \, \text{hours} \].This is the time after which you should be ready to hit the ball.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Constant
The gravitational constant, denoted as \( G \), is a key figure in astronomy and physics. It represents the force of attraction between two unit masses separated by a unit distance. The value of \( G \) is approximately \( 6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2 \). This constant is crucial in the calculation of gravitational forces between celestial bodies, such as moons, planets, and stars. To understand its importance, consider how it enables us to calculate the gravitational pull Deimos exerts on a baseball. Without \( G \), we couldn't determine how quickly a baseball has to be thrown to achieve orbit around Deimos.
Orbital Velocity
Orbital velocity is the speed required to keep an object in a stable orbit around a celestial body. For circular orbits, this velocity ensures the centrifugal force balances the gravitational pull. The formula is given by:
  • \( v = \sqrt{\frac{GM}{R}} \)
Where \( G \) is the gravitational constant, \( M \) is the mass of the body being orbited, and \( R \) is the radius from the center of the body to the orbiting object.
For our exercise on Deimos, substituting the known values yields an orbital velocity of approximately \( 5.6 \, \text{m/s} \). This means a baseball must be thrown at this speed to circle Deimos and return to you, assuming no other forces act on it.
Orbital Period
The orbital period is the time an object takes to make one complete revolution around another object. For a circular orbit, this can be computed using the formula:
  • \( T = \frac{2\pi R}{v} \)
Here, \( T \) is the orbital period, \( \pi \) is approximately 3.14159, \( R \) is the radius of the circular path, and \( v \) is the orbital velocity. Substituting into this formula for Deimos yields an approximate period of \( 6,734 \, \text{s} \), which converts to just over 1.87 hours.
This indicates how long it would take a baseball to complete its journey around Deimos and return to its starting point.
Circular Orbit
A circular orbit is a path where an orbiting object maintains a constant distance from the center of the body it's orbiting. This occurs when the velocity of the orbiting object results in a perfect balance between the centripetal and gravitational forces.
Unlike elliptical orbits, which have varying distances from the center, circular orbits are simpler to analyze and have uniform motion attributes.
In the problem with Deimos, achieving such a circular orbit necessitates a precise throwing speed to ensure the baseball returns to the pitcher's position. Understanding circular orbits aids in grasping how objects, like satellites, maintain stable paths around planets and moons.

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Most popular questions from this chapter

An \(8.00-\mathrm{kg}\) point mass and a 15.0 -kg point mass are held in place 50.0 \(\mathrm{cm}\) apart. A particle of mass \(m\) is released from a point between the two masses 20.0 \(\mathrm{cm}\) from the \(8.00-\mathrm{kg}\) mass along the line connecting the two fixed masses. Find the magnitude and direction of the acceleration of the particle.

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A uniform wire with mass \(M\) and length \(L\) is bent into a semicircle. Find the magnitude and direction of the gravitational force this wire exerts on a point with mass \(m\) placed at the center of curvature of the semicircle.

In March \(2006,\) two small satellites were discovered orbiting Pluto, one at a distance of \(48,000 \mathrm{km}\) and the other at \(64,000 \mathrm{km} .\) Pluto already was known to have a large satellite Charon, orbiting at \(19,600 \mathrm{km}\) with an orbital period of 6.39 days. Assuming that the satellites do not affect each other, find the orbital periods of the two small satellites without using the mass of Pluto.

There are two equations from which a change in the gravitational potential energy \(U\) of the system of a mass \(m\) and the earth can be calculated. One is \(U=m g y(\) Eq. 7.2\() .\) The other is \(U=-G m_{\mathrm{E}} m / r(\mathrm{Eq} .13 .9) .\) As shown in Section \(13.3,\) the first equation is correct only if the gravitational force is a constant over the change in height \(\Delta y .\) The second is always correct. Actually, the gravitational force is never exactly constant over any change in height, but if the variation is small, we can ignore it. Consider the difference in \(U\) between a mass at the earth's surface and a distance \(h\) above it using both equations, and find the value of \(h\) for which Eq. \((7.2)\) is in error by 1\(\% .\) Express this value of \(h\) as a fraction of the earth's radius, and also obtain a numerical value for it.
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